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23h
comment A bit of history of Verdier duality
This doesn't answer the question asked. Verdier himself certainly didn't formulate Verdier duality as an equivalence between categories of sheaves and cosheaves.
Dec
15
comment The cohomology of an $S_{3}$ cover of an elliptic curve ramified in one point
Already the easier question when C is a genus two curve covering an elliptic curve is nontrivial.
Dec
11
comment Example to show that the inverse image under a finite morphism is not t-exact with respect to the perverse t-structure
You're absolutely right, I managed to confuse myself when writing the answer. I changed the example, now it should be OK.
Dec
11
revised Example to show that the inverse image under a finite morphism is not t-exact with respect to the perverse t-structure
added 295 characters in body
Dec
11
revised Example to show that the inverse image under a finite morphism is not t-exact with respect to the perverse t-structure
added 26 characters in body
Dec
11
answered Example to show that the inverse image under a finite morphism is not t-exact with respect to the perverse t-structure
Dec
5
awarded  Nice Question
Dec
4
reviewed Approve Euler's Triangular Number closure properties
Dec
1
comment Automorphism group of a modular curve and its action on the set of cusps
A trivial remark. When $\Gamma = \Gamma(N)$ the automorphism group contains the quotient $\Gamma(1)/\Gamma(N) = \mathrm{SL}(2,\mathbf Z/N)$. It acts transitively on the cusps, since the quotient by the action is $X(1)$, which only has one cusp.
Nov
23
comment cohomology of a variation of wreath product
Consider the maps $(z_1,\ldots,z_n,z_{\sigma(1)}, \ldots,z_{\sigma(n)}) \mapsto (tz_1,\ldots,tz_n,tz_{\sigma(1)}, \ldots,tz_{\sigma(n)})$ for $t \in [0,1]$: this is a homotopy between the identity map and the constant map $(0,0,\ldots,0)$.
Nov
22
comment cohomology of a variation of wreath product
Are you sure you're asking the question you intended to ask? The space you wrote down is obviously contractible. You could insist that all $z_i$ are distinct but in this case you would obtain $n!$ copies of the usual configuration space of $n$ ordered points in $\mathbf C$.
Nov
18
awarded  Nice Answer
Nov
16
comment Cofree Lie Coalgebra
Also, your cofree Lie coalgebras should really be cofree conilpotent Lie coalgebras, to be pedantic.
Nov
16
comment Cofree Lie Coalgebra
The cofreeness is also clear in this picture. A tree as above looks like it describes an $n$-fold iterated comultiplication, so "apply" this tree to an element of $C$ to get an element of $A^{\otimes n}$ which is only well defined modulo the antisymmetry and Jacobi relations.
Nov
16
comment Cofree Lie Coalgebra
I don't have time to leave a detailed answer but this is explained in the book "Algebraic operads" by Loday-Vallette. In the first part of the book they explain that the tensor algebra quotiented by nontrivial shuffle products is naturally dual to the space of formal Lie words. You can think of $Lie^c(A) = Lie^c(n) \otimes_{S_n} A^{\otimes n}$ where $Lie^c(n)$ is the $S_n$-module spanned by binary trees with a single root and $n$ leaves, modulo the antisymmetry and Jacobi relations. The coproduct is given by a sum over all ways of splitting trees in half by deleting an edge.
Nov
16
awarded  Good Answer
Nov
13
awarded  Enlightened
Nov
13
awarded  Nice Answer
Nov
12
comment Framed version of braided monoidal category
You're right, they form a nonsymmetric operad. By the way, I think there are some things you might find useful in Nathalie Wahl's Ph.D. thesis.
Nov
12
comment Gerbes and Stacks
I don't really understand you so at least one of us is confused. The fiber over a point of $U_i$ is a groupoid with one object and morphism set $BGL(1)$.